|
|
|
|
| 0 | 1 | 0xa7 |
| 1 | 1 | 0xf1 |
| 2 | 1 | 0xd9 |
| 3 | 1 | 0x2a |
| 4 | 0 | 0x82 |
| 5 | 1 | 0xc8 |
| 6 | 1 | 0xd8 |
| 7 | 1 | 0xfe |
| 8 | 1 | 0x43 |
| 9 | 0 | 0x4d |
| 10 | 1 | 0x98 |
| 11 | 1 | 0x55 |
| 12 | 0 | 0x8c |
| 13 | 1 | 0xe2 |
| 14 | 1 | 0xb3 |
| 15 | 1 | 0x47 |
0x47f4 _____________ 0x4d9a _____________ 0x1a45 _____________ 0xa702 _____________ 0x1a42 _____________ 0x1b40 _____________ 0x1a45 _____________
Consider the following speculative Tomasulo's algorithm table. It is the solution to the same kind of problem you had on your homework, but for a machine with a slightly different MIPS configuration:

Answer the following questions about this machine based on information you derive from the table. Each answer should be short and fit in the space provided.